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kenny6666 [7]
3 years ago
9

Determine the average velocity and average speed for the following situation. A dog runs west for 35 meters then turns east for

45 meters stops then continues east for 65 meters. The time for the total trip is 85 seconds.
Physics
1 answer:
mart [117]3 years ago
7 0

Answer:

0.882 m/s average velocity and 1.71 m/s average speed

Explanation:

The dog travels a total of 35 m west and 110 m east.

110-35 = 75 m east of the starting position. Since velocity is a vector you must consider its first and final position and not the total distance traveled.

75 m / 85 s = 0.882 m/s average velocity

Speed is not concerned with direction so we instead add the total distance traveled which is 35+110 = 145 m. We then perform the same operation as before and divide by the time it took to run this distance.

145 m / 85 s = 1.71 m/s average speed

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An electron is a negatively charged particle that has a charge of magnitude, e - 1.60 x 10-19 C. Which one of the following stat
Ivahew [28]

Answer:

The electric field is directed toward the electron and has a magnitude of E=\frac{ke}{r^{2} }.

Explanation:

An electric field is define as the surrounding of charges which exert a force on each other and this force can be attractive or repulsive depends on the charge.

In the given case electron is given and the magnitude of charge on electron is e=1.6\times 10^{-19}C

Electric field can be represented as,

E=\frac{kQ}{r^{2} }

Here, r is the distance between the point ande charge, k is the electric field constant and Q is the charge.

In the given question an electron is given so electric field will be,

E=\frac{ke}{r^{2} }

As we know that electric field start from the positive charge and vanish in the negative charge.

So, here the electric field will be E=\frac{ke}{r^{2} } and it is directed toward the electron because of negative charge on the electron.

6 0
3 years ago
How much energy (in Joules) is released when 12.0 g of water cools from 20.0 °C to 11.0 °C? This is a grade 10 question from the
KATRIN_1 [288]

Answer: - 452.088joule

Explanation:

Given the following :

Mass of water = 12g

Change in temperature(Dt) = (11 - 20)°C = - 9°C

Specific heats capacity of water(c) = 4.186j/g°C

Q = mcDt

Where Q = quantity of heat

Q = 12g × 4.186j/g°C × - 9°C

Q = - 452.088joule

7 0
3 years ago
A heavy object and a light object are dropped from the same height. If we neglect air resistance, which will hit the ground firs
Maksim231197 [3]

Answer:

None, both objects will hit ground at the same time.

Explanation:

  • Assuming no air resistance present, and that both objects start from rest, we can apply the following kinematic equation for the vertical displacement:

        \Delta h = \frac{1}{2}*g*t^{2}  (1)

  • As the left side in (1) is the same for both objects, the right side will be the same also.
  • Since g is constant close to the surface of the Earth, it's also the same for both objects.
  • So, the time t must be the same for both objects also.
6 0
3 years ago
The Hubble Space Telescope (HST) orbits 569,000m above Earth’s surface. Given that Earth’s mass is 5.97 × 10^24 kg and its radiu
Soloha48 [4]
Refer to the diagram shown below.

M = 5.97 x 10²⁴ kg, mass of the earth
h = 5.69 x 10⁵ m, height of HST above the earth's surface
R = 6.38 x 10⁶ m, radius of the earth

Note that
G = 6.67 x 10⁻¹¹ (N-m²)/kg², gravitational acceleration constant.
R + h = 6.38 x 10⁶ + 5.69 x 10⁵ = 6.949 x 10⁶ m

The force between the earth and HST is
F = (GMm)/(R+h)²

Let v = tangential velocity of the HST.

The centripetal force acting on HST is equal to F.
Therefore
m*[v²/(R+h)] = (GMm)(R+h)²
v² = (GM)/(R+h)
    = [(6.67 x 10⁻¹¹ (N-m²)/kg²)*(5.97 x 10²⁴ kg)]/(6.949 x 10⁶ m)
    = 5.7303 x 10⁷ (m/s)²
v = 7.5699 x 103 m/s

Answer
The tangential speed of HST is about 7,570 m/s 

3 0
4 years ago
A person is pushing 108 kg box with a 325 n force if the friction force is 55 n what is the acceleration of the box
kompoz [17]
The answer is 3.52 m/s^2
4 0
4 years ago
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